Question

If $\int \frac{3e^x-5e^{-x}}{4e^x+5e^{-x}}dx=ax+b\ln (4e^x+5e^{-x})+c$, then

• A. $a=-\frac{1}{8},b=\frac{7}{8}$
• B. $a=\frac{1}{8},b=\frac{7}{8}$
• C. $a=-\frac{1}{8},b=-\frac{7}{8}$
• D. $a=\frac{1}{8},b=-\frac{7}{8}$

Answer 1

The correct option is A $a=-\frac{1}{8},b=\frac{7}{8}$

For solving integrals of this type, we express our numerator as sums of denominator and its derivative.
i.e. For integral $I=\int \frac{N_t}{D_t}dx$,
we write $N_t=A\left(D_t\right)+B\left({D_t}^{\prime }\right)$
where ${D_t}^{\prime }$ is the derivative of denominator and $A\mathrm{and}B$ are constants.
Thus the integral becomes
$I=A\int \frac{D_t}{D_t}dx+B\int \frac{{D_t}^{\prime }}{D_t}dx$
$\Rightarrow I=Ax+B\ln \left(D_t\right)+c,$
c being constant of integration.

Now, for our integral
$I=\int \frac{3e^x-5e^{-x}}{4e^x+5e^{-x}}dx$
$\mathrm{we}\mathrm{write},3e^x-5e^{-x}=a(4e^x+5e^{-x})+b(4e^x-5e^{-x})$
Now, comparing the co-efficients of $e^x\mathrm{and}{e}^{-x}$, we get the following equations:
$a+b=\frac{3}{4}$, and $a-b=-1$,
Now solving these two equations for $a\mathrm{and}b$, we get $a=-\frac{1}{8},\mathrm{and}b=\frac{7}{8}$,
Thus Option a. is correct.